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Re: Quasi-fields
Posted:
Jul 18, 2011 8:13 AM
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On Mon, 18 Jul 2011, William Elliot wrote:
> On Mon, 18 Jul 2011, David Hartley wrote:
>>>>>>>
>>>>>>>> In Dauben's biography of Abraham Robinson (page 103 of the
>>>>>>>> first printing) a quasi-field is said to be a commutative
>>>>>>>> field with distribution
>>>>>>>>
>>>>>>>> a(b + c) = ab + ac (1)
>>>>>>>>
>>>>>>>> replaced by
>>>>>>>> a(b_1 + b_2 + ... + b_n) = ab_1 + ab_2 + ... + ab_n. (2)
>>>
>>>>> Thus, in a quasi-field, the law x*0 = 0 holds iff the
>>>>> distributive law holds.
>>>>>
>>> Define as usual for n in N, na = sum(j=1,n) a.
>>>
>>> a0 = a * n0 = n(a0); (n-1)(a0) = 0
>>>
>>> Thus quasi distributivity (QD).
>>> a(b + c) = a(b + c + (n-2)0) = ab + ac + (n-2)(a0) = ab + ac - a0
>>>
>>> Does (-1)a = -a hold in a quasi field?
>>
>> It requires proof. It does hold for n=3:
>>
>> a*b = a*(b + b + -b)
>> = a*b + a*b + a*(-b)
>>
>> so a*b + a*(-b) = 0, a*(-b) = -(a*b)
>>
Using QD
a0 = a(1 - 1) = a1 + a(-1) - a0
a(-1) = a0 + a0 - a
If n = 3:
a0 = a(0 + 0 + 0) = a0 + a0 + a0
a0 + a0 = 0
a(-1) = -a
> In general for n = 2k + 1:
>
> a0 = a(kb + k(-b) + 0) = k(ab) + k(a(-b) + a0
>
> k(a(-b)) = -k(ab)
>
>> But that's no help for n even.
>
> For n = 2k:
> a0 = a(kb + k(-b)) = k(ab) + k(a(-b))
>
> k(a(-b)) = a0 - k(ab)
>
> Maybe a0 is an infinitesimal.
>
>>> Assume it does.
>>> a0 = a(1 - 1) = a1 + a(-1) - a0 = a - a - a0
>>> a0 + a0 = 0
>>> a0 = a * n0 = n(a0)
>>>
>>> If n is even, then a0 = 0 and we've a field.
Misc.
(-1)0 = 0*0 + 0*0
(1 + 1)a = a1 + a1 - a0 = a + a - a0
(1 + 1)0 = -(0*0)
>
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